Subject: Chemistry — Electrochemistry
Question: Consider a 70% efficient hydrogen-oxygen fuel cell
working under standard conditions at 1 bar and 298 K. Its cell reaction
is
$$H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l)$$
The work derived from the cell on the consumption of
$1.0 \times 10^{- 3}mol$ of $H_{2}(g)$ is used to compress $1.00\ mol$
of a monoatomic ideal gas in a thermally insulated container. What is
the change in the temperature (in K) of the ideal gas?
The standard reduction potentials for the two half-cells are given
below.
$$O_{2}(g) + 4H^{+}(aq) + 4e^{-} \rightarrow 2H_{2}O(l),E{^\circ} = 1.23V,$$
$$2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g),E{^\circ} = 0.00V$$
Answer: 13.32
Solution: For give reaction
$H_{2}(g) + \frac{1}{2}O_{2}(g)\overset{\ \ \ 2e^{-}\ \ \ }{\rightarrow}H_{2}O(l)$
$E{^\circ} = 1.23V$
$$\mathrm{\Delta}G{^\circ} = - FE{^\circ}_{cell} = \lbrack - 2 \times 96500 \times 1.23\rbrack 1 \times 10^{- 3} \times 0.7 = - 166.173\ J$$
$$W = \frac{nR\mathrm{\Delta}T}{\gamma - 1}$$
$$166.173 = \frac{1 \times 8.314 \times \mathrm{\Delta}T}{\left( \frac{5}{3} - 1 \right)} = \frac{8.314 \times 3}{2}\mathrm{\Delta}T$$
$$\mathrm{\Delta}T = \frac{166.173 \times 2}{8.314 \times 3} = 13.32$$